Scott's way of handling cardinals as objects without Choice depends on
foundation.
A couple of people have reported suggestions (proofs that are not
readily accessible for checking) that it CAN'T be done in ZF without
foundation.
I had for a long time THOUGHT that you could do something like this:
(i) we don't have RANKS, but if we had something that was a close
enough analogue of ranks to put a limit on the size of the candidates, we
could do something analogous to Scott's trick: take the cardinal to be the
set of candidates of some small size.
(ii) ZF without foundation lets you prove that the transitive closure
of a set (the set containing its members, its members's members, and so on)
exists. (I think this is useful in working with Aczel's Anti-Founded set
theory.) So, take as measure of the "size" of a set, the cardinality of
its transitive closure.
(iii) **CONJECTURE** At least in Aczel's ZF-AFA (though ??maybe not in
plain ZF with neither foundation nor anti-foundation??) I think (hope?) you
can prove that for any set there is a SET of all those sets whose
transitive closures are no larger than that of the given set. Call this
thing-- if it exists-- the "pseudorank" of the given set.
(iv) So-- no proof, and a poor track record with conjectures, to back
this up-- can't you, at least in Aczel's ZF-AFA, identify the cardinal of a
set with the set of all sets equal to it in cardinality and belonging to
its pseudorank?
(v) I guess this is consistent with the claim Kanovei and Matthias
report, that no definition works in plain ZF. If my conjecture in (iii) is
true, it probably DOES depend on AFA (which implies that every set has a
representing graph, which can be taken to be in the well-founded sets).
I'm puzzled, though.
---
Allen Hazen
Philosophy Department
University of Melbourne